IChO 1980 · Theory problem 1

Photodissociation of chlorine and the quantum yield of the H2-Cl2 reaction

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The dissociation of (molecular) chlorine is an endothermic process, \(\Delta H = 243.6\ \mathrm{kJ\,mol^{-1}}\).

The dissociation can also be attained by the effect of light.

1.1 At what wavelength can the dissociating effect of light be expected?

1.2 Can this effect also be obtained with light whose wavelength is smaller or larger than the calculated critical wavelength?

1.3 What is the energy of the photon with the critical wavelength?

When light that can effect the chlorine dissociation is incident on a mixture of gaseous chlorine and hydrogen, hydrogen chloride is formed. The mixture is irradiated with a mercury UV-lamp (\(\lambda = 253.6\ \mathrm{nm}\)). The lamp has a power input of 10 W. An amount of 2 % of the energy supplied is absorbed by the gas mixture (in a 10 litre vessel). Within 2.5 seconds of irradiation 65 millimoles of \(\ce{HCl}\) are formed.

1.4 How large is the quantum yield (= the number of product molecules per absorbed photons)?

1.5 How can the value obtained be (qualitatively) explained? Describe the reaction mechanism.

Official English text of the problem.

Answers

  • 1.1 The critical wavelength is \(\boxed{\lambda_{\mathrm{crit}} = 491.1\ \mathrm{nm}}\), the maximum wavelength permitted by the dissociation-energy requirement.
  • 1.2 Shorter wavelengths can cause dissociation if absorbed. Longer wavelengths cannot supply sufficient energy for single-photon dissociation.
  • 1.3 \(\boxed{E_{\mathrm{crit}} = 4.045\times10^{-19}\ \mathrm{J}}\) per photon.
  • 1.4 \(\boxed{\Phi \approx 6.1\times10^4}\) hydrogen chloride ([H]Cl) molecules per absorbed photon.
  • 1.5 The reaction proceeds by a radical chain mechanism. Photodissociation initiates chains whose propagation repeatedly produces hydrogen chloride and regenerates the radical carriers. Radical recombination terminates the chains.

Solution

Use the standard constants \[ h=6.62607015\times10^{-34}\ \mathrm{J\,s},\qquad c=2.99792458\times10^8\ \mathrm{m\,s^{-1}}, \] \[ N_{\mathrm A}=6.02214076\times10^{23}\ \mathrm{mol^{-1}}. \]

The chemical species used below are:

Name Formula SMILES
Molecular chlorine \(\ce{Cl2}\) ClCl
Molecular hydrogen \(\ce{H2}\) [H][H]
Hydrogen chloride \(\ce{HCl}\) [H]Cl
Chlorine atom (chlorine radical) \(\ce{Cl^{\bullet}}\) [Cl]
Hydrogen atom (hydrogen radical) \(\ce{H^{\bullet}}\) [H]

1.1 — Critical wavelength

The dissociation is \[ \ce{Cl2(g) -> 2Cl^{\bullet}(g)}. \]

Taking the given molar dissociation enthalpy as the required dissociation energy, one absorbed photon must supply at least \[ E_{\mathrm{diss}}=\frac{\Delta H}{N_{\mathrm A}}. \]

Since a photon has energy \(E=hc/\lambda\), the threshold satisfies \[ \frac{hc}{\lambda_{\mathrm{crit}}} =\frac{\Delta H}{N_{\mathrm A}}. \]

Therefore, \[ \lambda_{\mathrm{crit}} =\frac{N_{\mathrm A}hc}{\Delta H} =\frac{N_{\mathrm A}hc} {243.6\times10^3\ \mathrm{J\,mol^{-1}}} =4.91078\times10^{-7}\ \mathrm m. \]

Thus, \[ \boxed{\lambda_{\mathrm{crit}}=491.1\ \mathrm{nm}}. \]

1.2 — Shorter and longer wavelengths

Photon energy is inversely proportional to wavelength: \[ E=\frac{hc}{\lambda}. \]

Consequently:

  • For \(\lambda<\lambda_{\mathrm{crit}}\), a photon has more than the minimum dissociation energy. Dissociation is energetically possible, provided the photon is absorbed in a suitable transition.
  • For \(\lambda>\lambda_{\mathrm{crit}}\), a photon has insufficient energy. It cannot dissociate the molecule in the single-photon process considered here.

1.3 — Energy of a critical-wavelength photon

\[ E_{\mathrm{crit}} =\frac{\Delta H}{N_{\mathrm A}} =\frac{243.6\times10^3\ \mathrm{J\,mol^{-1}}} {6.02214076\times10^{23}\ \mathrm{mol^{-1}}}. \]

Hence, \[ \boxed{E_{\mathrm{crit}}=4.045\times10^{-19}\ \mathrm J}. \]

1.4 — Quantum yield

The energy absorbed during irradiation is \[ Q_{\mathrm{abs}} =0.02Pt =0.02(10\ \mathrm{J\,s^{-1}})(2.5\ \mathrm s) =0.50\ \mathrm J. \]

At the lamp wavelength, the energy per photon is \[ E_{\mathrm{UV}} =\frac{hc}{253.6\times10^{-9}\ \mathrm m} =7.833\times10^{-19}\ \mathrm J. \]

The number of absorbed photons is therefore \[ N_{\mathrm{abs}} =\frac{Q_{\mathrm{abs}}}{E_{\mathrm{UV}}} =\frac{0.50}{7.833\times10^{-19}} =6.383\times10^{17}. \]

The number of hydrogen chloride molecules formed is \[ N_{\ce{HCl}} =n_{\ce{HCl}}N_{\mathrm A} =(65\times10^{-3}\ \mathrm{mol})N_{\mathrm A} =3.914\times10^{22}. \]

Using the definition in the question, \[ \Phi =\frac{N_{\ce{HCl}}}{N_{\mathrm{abs}}} =\frac{3.914\times10^{22}}{6.383\times10^{17}} =6.132\times10^4. \]

Thus, \[ \boxed{\Phi\approx6.1\times10^4 \ \text{hydrogen chloride molecules per absorbed photon}}. \]

The vessel volume is unnecessary because the calculation uses total absorbed energy and total product amount.

1.5 — Explanation and reaction mechanism

The very large quantum yield shows that one absorbed photon leads to many product-forming reactions through a radical chain.

Initiation: \[ \ce{Cl2 ->[h\nu] 2Cl^{\bullet}}. \]

Light causes homolytic cleavage of the chlorine–chlorine bond: one bonding electron goes to each chlorine atom.

Propagation: \[ \ce{Cl^{\bullet} + H2 -> HCl + H^{\bullet}}, \] \[ \ce{H^{\bullet} + Cl2 -> HCl + Cl^{\bullet}}. \]

First, a chlorine radical abstracts a hydrogen atom from molecular hydrogen. The resulting hydrogen radical then abstracts a chlorine atom from molecular chlorine.

In each abstraction, the incoming radical’s unpaired electron and one electron from the broken bond form the new bond; the other bond electron remains on the atom released as a radical. These electron movements are not depicted by the SMILES.

Adding the two propagation steps gives \[ \boxed{\ce{H2 + Cl2 -> 2HCl}}. \]

The chlorine radical is regenerated, so this cycle can repeat many times without another photon.

Termination: radicals are removed by recombination, for example, \[ \ce{Cl^{\bullet} + Cl^{\bullet} + M -> Cl2 + M}, \] \[ \ce{H^{\bullet} + Cl^{\bullet} + M -> HCl + M}, \] \[ \ce{H^{\bullet} + H^{\bullet} + M -> H2 + M}. \]

Here \(\ce{M}\) denotes a third collision partner that carries away excess energy. Radicals may also be lost at the vessel walls.

Because propagation occurs many times before termination, approximately \(6.1\times10^4\) hydrogen chloride molecules can be formed per absorbed photon. The photon initiates the chain; the overall exothermic chemical reaction supplies the energy released during its continuation.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.